Liquids and Solutions Crib Sheet Determining the melting point of a substance from its solubility Consider a saturated solution of B in a solvent, A. Since the solution is saturated, pure solid B is in equilibrium with the B in solution: µ B(s) = µ B(l) and so if B is present in solution at a mole fraction x B : i.e. µ B(s) = µ B(l) + T ln x B ln x B = µ B(s) µ B(l) = fusg B T T since the chemical potential of a pure substance is simply the molar Gibbs free energy. ecall that fus G B is the Gibbs free energy change for the process B (s) B (l). Using G = H T S, we have ln x B = fush B T + fuss B Now, at T = T, the melting point of B, fus G B = 0 and so we can add fus G B T to the right hand side to get ln x B = fush B T = fush B T + fuss B fuss B + ( fus H B T = 0 fuss B ) Assuming that fus H B and fuss B are constant across the temperature range of interest, we have ln x B = fush ( B 1 T 1 ) T Therefore, a plot of ln x B, the saturation mole fraction of B in A, against 1/T yields a straight line with slope fus H B / and intercept fush B /T. That is, we can deduce the melting point of B and its enthalpy of fusion from measurements of its solubility at different temperatures.
Determining the molar mass of a solute Consider a solution of B in a solvent, A. At the melting point of the solvent, µ A(s) = µ A(l) = µ A(l) + T ln x A where x A = 1 x B. As before, ln(1 x B ) = µ A(s) µ A(l) T = fusg A T = fush A T + fuss A Now, if x B = 0 then T = T, the melting point of pure A, and fus G A = 0, so as before fus G A T = fush A T fuss A = 0 And if, as before, we assume that fus H A and fuss A are constant with temperature we can add this equation to the right hand side to get ln(1 x B ) = fush A ( 1 T 1 ) T Finally, if B is only sparingly soluble in A, x B 1 and we can use the approximation ln(1 + y) y for small y; that is ln(1 x B ) x B. Also, note that 1 T 1 T = T T T T T T 2 when T T. Thus, x B = fush A T T 2 So from a measurement of how the freezing point of the solvent changes ( T ) when a known mass of B is dissolved in it, x B can be found (if fus H A is known). But n B = x B (n A + n B ) x B n A So for a known amount of solvent, n A, we know how many moles of B have been dissolved. If this corresponds to a known (measured) mass of B, m, the molar mass of B is simply m B = m/n B.
Mixing of ideal solutions (or gases) Consider two pure liquids, A and B, separated by a partition. Their total Gibbs free energy is = n A µ A + n B µ B G i where there are n A moles of A and n B moles of B - recall that the chemical potential of a pure substance is the molar Gibbs free energy, µ = G m. Suppose now that the partition is removed and the liquids mix. If the solution is ideal, there is no enthalpy change on mixing: mix H ideal = 0 In an ideal solution the A-B interaction energy is the same as the average A-A and B-B interaction energy and the driving force for mixing is purely entropic: A and B molecules are randomly distributed about one another. The total Gibbs free energy of the mixture is G f = n A (µ A + T ln x A ) + n B (µ B + T ln x B ) where x A is the mole fraction of A in the mixture and x B is the mole fraction of B. Therefore, the Gibbs free energy change on mixing is mix G ideal = G f G i = n A T ln x A + n B T ln x B Now, n A = x A n and n B = x B n, where n = n A + n B. Thus, Furthermore, since S = ( G/ T ) p, mix G ideal = nt (x A ln x A + x B ln x B ) mix S ideal = n(x A ln x A + x B ln x B ) egular solutions eal solutions have values for mix G, mix H, and mix S that differ from their ideal values. The difference is called the excess: mix G = mix G ideal + G excess mix H = H excess mix S = mix S ideal + S excess A regular solution is one for which there is a non-zero enthalpy of mixing, but no excess entropy of mixing (S excess = 0): the A-B interaction is different from the average A-A and B-B interaction, but not so much so that the A and B molecules are no longer randomly distributed. This suggests that the enthalpy of mixing should be defined as H excess = mix H = βx A x B where β = zn A ɛ, z is the co-ordination number of both A and B, and ɛ = ɛ AB 1 2 ɛ AA 1 2 ɛ BB is the energy change when one A-B bond is formed in the mixture.
Osmotic pressure and polymer solution Consider a semi-permeable membrane separating pure solvent, A, and a solvent containing a solute, B, which cannot pass through the membrane. The pure solvent is subject to a pressure, p (usually just the ambient atmospheric pressure), and the solution subject to a greater pressure, p + Π which prevents the influx of solvent across the membrane. At equilibrium, µ A(p) = µ A (p + Π, x A ) where x A is the mole fraction of solvent on the solution side of the membrane. Therefore, µ A(p) = µ A(p + Π) + T ln x A Now, for a pure substance, µ = G m, and dg = V dp SdT, so at constant temperature dµ = V m dp. Thus, µ A(p + Π) = µ A(p) + p+π p V m dp If V m is constant with p (the solvent is incompressible), this becomes and so substituting in above, µ A(p + Π) = µ A(p) + V m Π 0 = V m Π + T ln x A If x B = 1 x A 1 (the solute is only sparingly soluble), and we have ln x A = ln(1 x B ) x B Π = T x B V m But x B = n B /(n A + n B ) n B /n A when n A n B, so Π = T n B, since V = V m n A V The concentration of the solute, [B] = n B /V, so Π = T [B] This equation is valid for ideal solutions. Solutions of polymers (even dilute solutions) are not ideal, and it is usual to treat them with a virial expansion: Π [B] = T (1 + B [B] + ) where B is the second virial coefficient. This enables the molar mass of a solute to be obtained because [B] = c B /M B, where c B is the molality of B in
the solution (the known mass of B dissolved e.g. in g dm 3 ), and M B is the molar mass of B (e.g. in g mol 1 ), so Π c B = T M B (1 + Bc B ) A plot of Π/c B against c B should yield a straight line with slope T B/M B and intercept T/M B. Note that B = B /M B is temperature- and solvent-dependent. The so-called θ-condition occurs when B = 0 and the solution behaves ideally (cf. the Boyle temperature for real gases). Debye-Hückel theory Debye-Hückel theory describes the behaviour of dilute ionic solutions. Each ion is thought of as being a point charge surrounded by a diffuse ionic atmosphere of opposite charge (when the thermal motions of all the surrounding ions has been averaged) present in a neutral, inert solvent which is considered to be a continuous medium. Each ion therefore gives rise to a shielded Coulomb potential which decreases exponentially as a function of distance from the ion (of charge z i ): φ i (r) = z i r e r/rd where r D is the Debye length, and depends on the ionic strength, I: r D I 1/2, where I = 1 2 z 2 i c i c Here, c i is the concentration of ions of charge z i. A larger ionic strength gives rise to greater shielding. The chemical potential of the ions is given in terms of their activities, a i µ i = µ i + T ln a i where a i = γ i c i c and γ i is the activity coefficient; Debye-Hückel theory provides a way of estimating γ i for weak ionic solutions: log 10 γ ± = z + z A I where A is a constant which depends on the solvent and temperature (A = 0.509 for water at 298 K).